Question

It $y=\frac{\sqrt{1-\sin 4x}+1}{\sqrt{1+\sin 4x}-1}$ then one of the values of y is/are-

• A. $\cot x$
• B. $-\tan x$
• C. $-\cot (\frac{\pi }{4}+x)$
• D. $\tan (\frac{\pi }{4}+x)$

Answer 1

Answer: (A), (D)

The correct options are
A $\cot x$
D $\tan (\frac{\pi }{4}+x)$

$y=\frac{\sqrt{1-\sin 4x}+1}{\sqrt{1+\sin 4x}-1}$

$=\frac{\sqrt{{(\sin 2x-\cos 2x)}^2}+1}{\sqrt{{(\sin 2x+\cos 2x)}^2}-1}$

$\Rightarrow y=\frac{|\sin 2x-\cos 2x|+1}{\sin 2x+\cos 2x-1}$

Case I: $\sin 2x>\cos 2x$
$\Rightarrow y=\frac{\sin 2x+1-\cos 2x}{\sin 2x+\cos 2x-1}$

$=\frac{2\sin x\cos x+2{\sin }^{2}x}{2\sin x\cos x-2{\sin }^{2}x}$

$=\frac{\cos x+\sin x}{\cos x-\sin x}$

$=\frac{1+\tan x}{1-\tan x}$

$y=\tan (\frac{\pi }{4}+x)$

Case II, $\sin 2x<\cos 2x$
$\Rightarrow y=\frac{\cos 2x-\sin 2x+1}{\sin 2x+\cos 2x-1}$

$=\frac{2{\cos }^{2}x-2\sin x\cos x}{2\sin x\cos x-2{\sin }^{2}x}$

$=\frac{2\cos x(\cos x-\sin x)}{2\sin x(\cos x-\sin x)}$

$\Rightarrow y=\cot x$